In triangle ABC, K and L are points on BC. K is closer to B than L. AL is the bisector of angle KAC. KL times BC=BK times CL. Find angle BAL.

Conrad Beltran 2022-09-20 Answered
In Δ A B C ,, K and L are points on BC. AL is the bisector of K A C. K L × B C = B K × C L. Find B A L.
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Answers (2)

Julien Zuniga
Answered 2022-09-21 Author has 7 answers
Step 1
The identity K L B C = B K C L implies that line AB is the external bisector of triangle CKA at A while AL is the interior bisector at A.
Step 2
Hence they are orthogonal and thus the angle B A L = 90 .
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Hagman7v
Answered 2022-09-22 Author has 3 answers
Step 1
By the bisector theorem we have A C A K = C L K L = B C B K hence μ = A C C B = A K K B and K belongs to the Apollonius circle that is the locus of points P such that P A P B = λ. Such a circle goes through the feet of the angle bisectors (the internal and the external) from C. If ABC is an acute-angled triangle, such a circle intersects the BC segment only at C, hence there is no solution.

On the other hand, if we allow B A C ^ to be greater than 90 , we have a solution at B A L ^ = 90 :
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